Chapter21: Potentiometry
Section: Chapter Questions
Problem 21.9QAP
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Question
What is the pH of (i) 1.0 X 103 M HBr and (ii) 1.0 X 102 M KOH?
Expert Solution
Step 1
We know that, pH and pOH can be expressed as the negative logarithm of hydrogen ion and hydroxide ion respectively, i.e.
pH = -log[H+]
and, pOH = -log[OH-]
Also, pH + pOH = 14
Part (a)
We have, [HBr] = 1.0 x 10-3 M
Since, HBr is an strong electrolyte, hence it dissociate completely to give H+ and Br-.
Thus, [H+] = [HBr] = 1.0 x 10-3 M
So, pH = -log[H+]
= -log(1.0 x 10-3)
= 3
Therefore, the pH of the solution is 3.
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